To solve the equation:

\[
\tan(3\alpha + 12^\circ) = \cot(\alpha + 36^\circ)
\]

we can use the identity \(\cot(x) = \frac{1}{\tan(x)}\), which gives:

\[
\tan(3\alpha + 12^\circ) = \frac{1}{\tan(\alpha + 36^\circ)}
\]

Taking the reciprocal of both sides, we get:

\[
\tan(3\alpha + 12^\circ) \cdot \tan(\alpha + 36^\circ) = 1
\]

Now, we have:

\[
\tan(3\alpha + 12^\circ) \cdot \tan(\alpha + 36^\circ) = 1
\]

This is a standard trigonometric equation, and we will use it to solve for \(\alpha\). Let me work through the next steps to simplify and find the solution.

First, rewrite the tangent product identity:
We know that:

\[
\tan(A) \cdot \tan(B) = 1 \quad \text{implies} \quad A + B = 90^\circ \quad \text{or} \quad n\pi/2
\]

In our case:

\[
(3\alpha + 12^\circ) + (\alpha + 36^\circ) = 90^\circ
\]

Simplifying:

\[
4\alpha + 48^\circ = 90^\circ
\]

Subtracting 48° from both sides:

\[
4\alpha = 42^\circ
\]

Dividing by 4:

\[
\alpha = 10.5^\circ
\]

Thus, the solution is:

\[
\alpha = 10.5^\circ
\]

### Do you want more details or have any questions?

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Here are 5 related questions:
1. How would the solution change if we had \(\cot(3\alpha + 12^\circ)\) instead of \(\tan(3\alpha + 12^\circ)\)?
2. What happens when \(\alpha\) is negative in this kind of equation?
3. How can we express \(\tan(x)\) in terms of \(\sin(x)\) and \(\cos(x)\)?
4. What other methods could we use to solve \(\tan(A) \cdot \tan(B) = 1\)?
5. How would this change in radians instead of degrees?

**Tip**: In trigonometric equations, converting everything to a single function like \(\tan\) or \(\sin\) can make the equation easier to solve.

This step-by-step guide explains how to solve the trigonometric equation tan(3α + 12°) = cot(α + 36°) using reciprocal identities and standard formulas. Learn key trigonometric concepts like tangent, cotangent, and the reciprocal identity. Suitable for students in Grades 10-12.

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